torsional deflection formula help (not homework)

[grunt58]

First off this is not homework; I'm self-studying and Tooling U for CMfgT certification and am confused why the force for T1 in the attached problem is doubled.

T1 = 4000 in-lb

Thanks

 

[dgallup]

It says right in the problem definition that there are two (2) 2000 in-lb torsional loads applied.
The torsional loads are additive. 2000 + 2000 = 4000

The diagram is pretty poor. It really should show that the left hand end is fixed and that that is where T1 is.

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The Help for this program was created in Windows Help format, which depends on a feature that isn't included in this version of Windows.

 

[grunt58]

Ahhh ok got it now; right over my head. Studying for the cert (work is paying). I feel confident in the manufacturing side. The applied engineering I'm going to struggle with.

Thanks again.

 

[chicopee]

Wait a minute the problem in the exercise is different from your OP. The problem ask for the torsional deflection in the entire length, 340", of the two shafts. So to calculate it, assume the junction between the 2" and 3" shaft to be stationary, then calculate the torsional deflection at that junction for the 2" shaft, 100 in long, exposed to a torsion of 2000 in-lb.; then calculate the torsional deflection of the 3", 240 in long, shaft exposed to a torsion of 4000 in-lb. Add the two torsional deflection values in order to meet the result ask for a torsional deflection of a 340" long member. Don't forget the 2" shaft is hollow.

 

[rb1957]

as the worked answer shows, at the tip the torque is 2000 in.lbs, at the base 4,000 in.lbs.
theta3 is the twist from the end to the base of the tube (100in from the load),
theta2 is the twist of the outer solid rod (3" dia), 2000 in.lbs for a length of 140in,
theta1 is the twist of the inner solid rod (3" dia), 4000 in.lbs over a length of 100in.
twists all add up.

another day in paradise, or is paradise one day closer ?